AS Daily A Level Physics question
In a photoelectric experiment, a clean metal with threshold frequency f0 is illuminated either by Lamp X (single spectral line at 1.2 f0) or Lamp Y (single spectral line at 2.2 f0). A lens is used with Lamp X so that the intensity at the metal for X is twice that for Y. You measure the magnitude of the stopping potential Vs needed to just reduce the current to zero. Which statement must be true about Vs(Y)/Vs(X), and why?
Answer
The correct answer is C.
Correct: C — Vs(Y)/Vs(X) = 6, because only how far the frequency exceeds the threshold sets the electron energy, and intensity does not change it. The maximum electron energy is proportional to (f − f0), so Vs ∝ (f − f0); thus Vs(Y)/Vs(X) = (2.2f0 − f0)/(1.2f0 − f0) = 1.2/0.2 = 6, and changing intensity does not alter energy per electron. A The ratio 2 ignores the fixed energy needed to escape; if f is just above threshold, subtracting f0 makes the lower-frequency case much smaller, so the ratio is not about “twice.” B Confuses intensity with photon energy; higher intensity gives more electrons per second, not higher energy per electron, so Vs does not increase with intensity. C This is the correct reasoning and arithmetic, using the excess above threshold and noting intensity’s irrelevance to energy. D Takes Vs proportional to f alone (2.2/1.2 ≈ 11/6) and ignores the subtraction of the threshold; near threshold the lower-frequency case gives a much smaller Vs, making the ratio much larger than 1.8.